Let `vec(a)` , `vec(b)` and `vec(c)` be three non-zero vectors such that no two of them are collinear and `(vec(a)×vec(b))×vec(c)=1/3|vec(b)||vec(c)|vec(a)`. If `theta` is the angle between vectors `vec(b)` and `vec(c)`, then the value of `sintheta` is:

Options

A. `(1)/(3)`
B. `(sqrt(2))/(3)`
C. `(2)/(3)`
D. `(2sqrt(2))/(3)`

Answer

To solve the problem step by step, we will use the properties of vectors and the triple product.

**Given:**
\[ (\vec{a} \times \vec{b}) \times \vec{c} = \frac{1}{3} |\vec{b}| |\vec{c}| \vec{a} \]

**Step 1: Apply the triple product identity**
Using the vector triple product identity, we have:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \]
Let \(\vec{x} = \vec{a}\), \(\vec{y} = \vec{b}\), and \(\vec{z} = \vec{c}\). Thus,
\[ (\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \]

**Step 2: Set the equation**
From the given equation, we can equate:
\[ (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \frac{1}{3} |\vec{b}| |\vec{c}| \vec{a} \]

**Step 3: Rearranging the equation**
Rearranging gives us:
\[ (\vec{a} \cdot \vec{c}) \vec{b} = \frac{1}{3} |\vec{b}| |\vec{c}| \vec{a} + (\vec{a} \cdot \vec{b}) \vec{c} \]

**Step 4: Dot product considerations**
Taking the dot product of both sides with \(\vec{a}\):
\[ \vec{a} \cdot \left((\vec{a} \cdot \vec{c}) \vec{b}\right) = \vec{a} \cdot \left(\frac{1}{3} |\vec{b}| |\vec{c}| \vec{a} + (\vec{a} \cdot \vec{b}) \vec{c}\right) \]
This simplifies to:
\[ (\vec{a} \cdot \vec{c})(\vec{a} \cdot \vec{b}) = \frac{1}{3} |\vec{b}| |\vec{c}| |\vec{a}|^2 + (\vec{a} \cdot \vec{b})(\vec{a} \cdot \vec{c}) \]

**Step 5: Canceling terms**
Now, we can cancel the \((\vec{a} \cdot \vec{c})(\vec{a} \cdot \vec{b})\) from both sides:
\[ 0 = \frac{1}{3} |\vec{b}| |\vec{c}| |\vec{a}|^2 \]
This implies that:
\[ \vec{a} \cdot \vec{c} = 0 \]
Thus, \(\vec{a}\) and \(\vec{c}\) are orthogonal.

**Step 6: Finding \(\cos \theta\)**
Next, we know that:
\[ \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \theta \]
From our earlier steps, we can substitute:
\[ - \vec{b} \cdot \vec{c} = \frac{1}{3} |\vec{b}| |\vec{c}| |\vec{a}| \]
Thus:
\[ \cos \theta = -\frac{1}{3} \]

**Step 7: Finding \(\sin \theta\)**
Using the identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting \(\cos \theta\):
\[ \sin^2 \theta + \left(-\frac{1}{3}\right)^2 = 1 \]
\[ \sin^2 \theta + \frac{1}{9} = 1 \]
\[ \sin^2 \theta = 1 - \frac{1}{9} = \frac{8}{9} \]
Thus:
\[ \sin \theta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \]

**Final Answer:**
\[ \sin \theta = \frac{2\sqrt{2}}{3} \]